2 Answers
2

Your reasoning is incorrect. It is not true that we have four spaces in AII to fit two characters. You seem to be assuming that C and TTT cannot be consecutive, which is not the case. Notice that ACTTTII is a valid arrangement of the letters before the S's are inserted.

Here is another method of thinking about the problem. If we set aside the three S's for the moment, we have five objects to arrange. They are an A, a C, two I's, and a block consisting of three T's. We can choose two of the five positions for the I's in $\binom{5}{2}$ ways, then arrange the remaining three objects in $3!$ ways. However, by symmetry, in only one third of these arrangements does the A precede both I's. Hence, the number of permissible ways of arranging these objects is
$$\frac{1}{3}\binom{5}{2}3! = \binom{5}{2}2!$$
To ensure no two S's are consecutive, we must choose three of the six available spaces (four between successive objects in a permissible arrangement of A, C, I, I, and TTT and the two ends of the row) in which to insert an S, which can be done in $\binom{6}{3}$ ways. Hence, the number of arrangements that satisfy the requirements is, in fact,
$$\binom{5}{2}2!\binom{6}{3}$$

$\begingroup$Ah! Thank you for pointing out my invalid assumption. It also made sense for me to think about the problem as ${4 \choose 1}{5 \choose 1}{6 \choose 3}$ which is also the same as ${5 \choose 2}2!{6 \choose 3}$.$\endgroup$
– James TaylorDec 11 '16 at 1:54